Factorising x2-100: As there is no common element in x2 or 100, we know this must consist of two sets of brackets not just one. we know both brackets must contain and x, and that which ever two numbers multiply to make 100 should be equal, cancelling out any number of x's as there are none in the equation. Therefore it could not be (x+50)(x-2) as this would leave you with x2+48x -100. It could be 10 in each bracket as 100 is a square number, giving you (x+10)(x-10). If we expand this out we get x2+10x -10x - 100. The two 10x's cancel out giving us x2-100, as required!